Maths ( Problems in 2 dimensions ) ! Urgent !!

Let PQ extents along the beach, there exist a point C to the east of Q where BC represents the shortest distance of buoy B from the beach.

Then BC / CQ = tan (90 - 20) = tan 70
BC = CQ tan 70 -- (1)

BC / CP = tan (90 - 50) = tan 40
BC / (CQ + QP) = tan 40
BC / (CQ + 200) = tan 40
BC = (CQ + 200) tan 40 -- (2)

Solving (1) & (2): get CQ = ??? -- (3)

Put (3) into (1): get BC!!

My answer: 87.9m

**********************************buoy B******
********************************-***-****_******
****************************-******-*****_******
************************-*********-******_******
********************-************-*******_******
****************-***************-********_******
************-******************-*********_******
**********P-------200m-------Q*********_R****


角BPQ
=90度 - 50度
=40度

角BQP
=90度 + 20度
=110度

看見有個直角三角形沒有？
設那個直角為R.

角BQR
=90度 - 20度
=70度

the distance of a buoy B from the beach
= the distance between buoy B and R

let the distance between buoy B and R be BR.
tangent 角BPQ = BR over PR
tan 40 = BR over PR
PR = BR over tan 40

tangent 角BQR = BR over QR
tan 70 = BR over QR
QR = BR over tan 70

因此 不用求PR和QR 就能找到BR

PR - QR =200m
(BR over tan 40) - (BR over tan 70) = 200m
BR = 200m over [ (1 over tan 40) - (1 over tan 70) ]

since tan X = 1 over tan (90-X)

BR = 200m over [ (1 over tan 40) - (1 over tan 70) ]
BR = 200m over [ tan 50 - tan 30 ]
BR = 325.52m (corrected to 2 decimal places)

清楚嗎？有問題再給我留言。

let the distant be x
x/tan20 + x/tan50 = 200
(tan20+tan50)x/tan20 tan50=200
(tan20+tan50)x=86.75256
x=55.76
the distant is 55.76m